Question

if alpha, beta and gamma are the roots of
the equation x^3 - 4x^2+6X-4 = 0.
the value of 1/alpha + 1/beta + 1/gamma is .



​

Answer 1

Answer:

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The value of ( 1/alpha + beta) + (1/beta + gamma) + (1/alpha + gamma) is 2

Solution:

Given that, f(x)=x^{3}+4 x+2f(x)=x

3

+4x+2

On comparing it with general Equation a x^{3}+b x^{2}+c x+dax

3

+bx

2

+cx+d we get,

a = 1, b = 0 , c = 4 and d = 2

\text {Also, } \alpha, \beta, \gamma \text {are the zeroes of the given cubic polynomial }Also, α,β,γare the zeroes of the given cubic polynomial

\text {Sum of zeroes of cubic polynomial }=\frac{-b}{a}Sum of zeroes of cubic polynomial =

a

−b

\alpha+\beta+\gamma = \frac{-0}{1}=0α+β+γ=

1

−0

=0 ---- eqn 1

\text {Sum of product of two zeroes of cubic polynomial }=\frac{c}{a}Sum of product of two zeroes of cubic polynomial =

a

c

\alpha \times \beta+\beta \times \gamma+\gamma \times \alpha=\frac{4}{1}=4α×β+β×γ+γ×α=

1

4

=4

\text {Product of zeroes of cubic polynomial }=\frac{-d}{a}Product of zeroes of cubic polynomial =

a

−d

\alpha \times \beta \times \gamma=\frac{-2}{1}=-2α×β×γ=

1

−2

=−2

We need to find value of :-

\frac{1}{\alpha+\beta}+\frac{1}{\beta+\gamma}+\frac{1}{\gamma+\alpha}

α+β

1

+

β+γ

1

+

γ+α

1

From equation (1), we get

=\frac{1}{-\gamma}+\frac{1}{-\alpha}+\frac{1}{-\beta}=

−γ

1

+

−α

1

+

−β

1

=\frac{-(\alpha \times \beta+\beta \times \gamma+\gamma \times \alpha)}{\alpha \times \beta \times \gamma}=

α×β×γ

−(α×β+β×γ+γ×α)

Subsituting the values we get,

=\frac{-(\alpha \times \beta+\beta \times \gamma+\gamma \times \alpha)}{-2}=

−2

−(α×β+β×γ+γ×α)

=\frac{-(4)}{-2}=2=

−2

−(4)

=2

Hence the value is found as 2

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Answer 2

What would be the denominator after rationalizing 7/ (5√3 - 5√2) ? *

1 point

a) 19

b) 20

c) 25

d) None of these